Question

Use the limit definition to find the derivitive of f(x)=1/(square root of x+1)

Answer 1

\[f(x) = \frac{1}{\sqrt{x+1}}\]

Answer 2

lim as x aproaches 0 of \[1\div \sqrt{x+1}\]

Answer 3

Is the limit Definition like this
\[\lim_{h \to 0} \:\:\:\:\:f(x+h) -f(x)\:\:?\]

Answer 4

I am confused about it, If you can tell me a little about it then I might be able to help you.

Answer 5

Use the limit definition above to solve for the derivative of f(x)

Answer 6

\[f(x +h ) = \frac{1}{\sqrt{x +h + 1 } }\]
\[\lim_{h \to 0 } \frac{1}{\sqrt{x+h+1}} - \frac{1}{\sqrt{x+1}}\]
\[\lim_{h \to 0 } \frac{\sqrt{x+1} - \sqrt{x+h+1}}{\sqrt{x+h+1}\times\sqrt{x+1}} \]

Answer 7

Now Rationalization of Numerator...

Answer 8

\[\lim_{h \to 0 } \frac{x +1 - x -h -1 }{ \sqrt{x+h+1}\times{\sqrt{x+1}(\sqrt{x+1} + {\sqrt{x+h+1}})}} \]

Answer 9

definition of the derivative is all over h noob

Answer 10

\[\lim_{h \to 0 } \frac{-h }{ \sqrt{x+h+1}\times{\sqrt{x+1}(\sqrt{x+1} + {\sqrt{x+h+1}})}}\]

Answer 11

should it all be over h so the h's cancel out?

Answer 12

oh sorry I told you I am confused about it sorry
yea h cancels out

Answer 13

boob

Answer 14

did I do something wrong now?

Answer 15

no, i think youre doing everything right, dont listen to that "The_mathster"

Answer 16

\[\lim_{h \to 0 } \frac{-1 }{ \sqrt{x+h+1}\times{\sqrt{x+1}(\sqrt{x+1} + {\sqrt{x+h+1}})}}\]h cancels out...And we have..
\[-\frac{1}{x +1 (2\sqrt{x+1})}\]
\[-\frac{1}{2}\times \frac{1}{(x+1)^{\frac{3}{2}}}\]

Answer 17

Limit complicates it, Differentiating is much easier...

Answer 18

yah it is, but our assignment asks for us to use limit definition. thanks for your help